CHAPTER 3

FORCE AND PRESSURE(textbook exercise)

Q1. Define Force. State its S.I. unit.

Force is that cause which changesthe state of a

body(either the state of rest or state of motion) or

changes the shape or size of the body.

Q2.State two effects of force when applied on a body.

Effects of Force:

1) A force when applied as push or pull on a stationary

body which is free to move, can produce motion in it

2) When force is applied as stretch or squeeze on a

body which is not free to move, it changes the shape

or size of the body.

Q3.How does the effect of a force differ when it is applied

on a (a) rigid body (b) non-rigid body.

(a) A force when applied on a rigid body can cause

only change in motion of the body.

(b) A force when applied on a non-rigid body can

cause both change in its size or shape and motion

in it.

Q5. Define the term moment of force.

Themoment of force is equal to the product of the

magnitude of the force and the perpendicular distance of

the force from the pivoted point.

Q6. State the S.I. unit of moment of force.

Unit of moment of force

= unit of force x unit of distance

The S.I. unit of force is newton and that of distance is

metre, so the S.I. unit of moment of force is newton x

metre which is written in short form as Nm

Q7. State the two factors which affect the moment of

force.

The moment of force on a body depends on:

(a) The magnitude of the force applied.

Larger the magnitude of force applied, more is the

moment of force.

(b) The perpendicular distance of the force from the

pivoted point.

Larger the perpendicular distance of point at

which the force is applied, from the pivoted point,

more is the moment of force

Q9. Write the expression for the moment of force about a

given axis of rotation.

In the figure given above, the perpendicular distance of

force F from the pivoted point O is OP. Therefore

Moment of force about the point O

= Force x perpendicular distance of force

from the point O

= F x OP

Q10.State one way to decrease the moment of a given

force about a given axis of rotation.

To decrease the moment of a given force about a given

axis of rotation, either decrease the force applied or

decrease the perpendicular distance of force from the

axis of rotation.

Q11.State one way to obtain greater moment of a given

force about a given axis of rotation.

For producing greater moment of force about a given

axis of rotation, the force should be applied on the body

at a point for which the perpendicular distance of the

force from the pivoted point is maximum so that the

given force may provide the maximum turning effect to

turn the body.

Q12. What do you mean by clockwise and anti-clockwise

moment of a force?

• Conventionally, if the effect on the body is to turn it

anticlockwise, moment of force is called

ANTICLOCKWISE MOMENT and it is taken as

POSITIVE.

• If the effect on the body is to turn it clockwise,

the moment of force is called as CLOCKWISE

MOMENT and taken as NEGATIVE.

Q13. Explain the following

(a) The spanner or wrench has long handle.

A spanner used to tighten or loosen a nut has a

long handle to produce a large turning effect by a

small force applied at the end of its handle as

shown in the figure below.

(b) The steering wheel of a vehicle is of large

diameter.

The steering wheel of vehicles can be turned by

applying a force at a point on the rim of the wheel.

If the diameter of the steering wheel will be large,

then the distance between the point of application

of force and the centre will increase. This will

produce the required turning effect by applying a

smaller force. Thus the driver can turn the steering

wheel easily as he will apply less force.

(c) The hand-flour grinder is provided with a handle

near the rim.

The hand flour grinder is provided with a handle

near its rim (i.e. at a maximum distance from the

centre) so that it can easily be rotated about the

iron pivot at its centre by applying a small force at

the handle.

(d) It is easier to open a door by pushing it at its free

end.

To open or shut a door we apply a force (push or

pull) perpendicular to the door at its handle. The

handle is provided at the maximum distance from

the hinges near the free end of the door so that a

smaller force at larger perpendicular distance

produces the required turning effect of force to

open or shut the door.

(e) A potter turns his wheel by applying a force

through the stick near the rim.

A potter’s wheel has a wheel pivoted at the centre.

The potter turns the wheel by means of a stick at

the rim of the wheel. This is done so as to increase

the distance between the point of application of

force from the pivoted point and thus the potter

can easily rotate the wheel by applying less force.

Q14. What is thrust?

The force acting normally on a surface is called thrust. A

body when placed on a surface exerts a thrust on the

surface equal to its own weight.

Q15. State the unit of thrust.

The unit of thrust is same as that of the weight or force.

Thus, the units of thrust are kilogram force (kgf), gram

force (gf) and newton (N). They are related as

1 kgf = 1000 gf

1 kgf = 10 N (nearly)

1 N = 100 gf (nearly)

Q16. On what factors does the effect of thrust on a

surface depend?

The effect of thrust on a surface depends on the area of

the surface on which it acts. Smaller the area of the

surface on which a thrust acts, larger is its effect. But the

effect of a thrust is less on a larger area.

Q17. Define the term ‘pressure’ and state its unit.

Pressure is defined as the thrust per unit area. Thus,

Pressure = 𝑇ℎ𝑟𝑢𝑠𝑡

𝐴𝑟𝑒𝑎

It is denoted by the letter P.

If a thrust F acts on an area A then pressure P =

P = 𝐹

𝐴

Units of Pressure

S.I. unit of Force Is newton (N) and that of area is metre2.

So the S.I. unit of pressure is newton/metre2 written as

N/m2 or Nm-2. When force is measured in kgf and area is

measured in cm2, pressureis expressed in unitkfg cm-2.

Q18. How is thrust related to pressure?

The effect of thrust is expressed in terms of a quantity

called pressure. More the effect of a given thrust on a

surface, it is said that thrust exerts more pressure on the

surface and if less is the effect of thrust on a surface, it is

said that the thrust exerts a less pressure on the surface.

Q19. Name two factors on which the pressure on a

surface depends.

The pressure on a surface depends:

1. On the area of the surface on which thrusts acts,

2. On the magnitude of thrust acting on the surface.

Q20. When does a man exert more pressure on the floor:

while standing or while walking?

When a man is walking, then at one time only his one

foot is on the ground. Due to this, the weight of man falls

on a smaller area of the ground and produces more

pressure on the ground. On the other hand, when the

man is standing, then both his feet are on the ground.

Due to this the weight of the man falls ona larger area of

the ground and produces lesser pressure on the ground.

Q21. Why camels and elephants have broad feet?

A camel has broad feet which reduces the pressure

exerted on the sand. As a result the camel’s feet do not

sink in the sand allowing it to move faster.

Elephants are very big in size. The broad feet of the

elephant distributes its weight on a larger area due to

which it exerts less pressure on the ground. This avoids

the elephant from sinking into the earth and it can walk

easily.

Q22.A sharp pin works better than a blunt pin. Explain

the reason.

A sharp pin has very small area of contact compared to a

blunt pin. On applying force the pointed end will exert

greater pressure and go deep into the given surface.This

is not possible in case of a blunt pin as area of contact is

more.

Q23. Why is the bottom part of the foundation of a

building made wide?

The bottom part of the foundation of a building is kept

wide so that the weight of the building may act on a

larger area. As a result, it will exert less pressure on the

ground. This avoids sinking of buildings into the earth.

Q24. It is easy to cut with a sharp knife than with a blunt

knife. Explain.

A sharp knife has a very thin edge to its blade. A sharp

knife cuts objects (like vegetables) better because due to

its very thin edge, the force of our hand falls over a very

small area of the object producing a large pressure and

this large pressure cuts the object easily. On the other

hand, a blunt knife has a thicker edge. A blunt knife does

not cut an object easily because due to its thicker edge,

the force of our hand falls over a larger area of the object

and produces lesser pressure. This lesser pressure cuts

the object with difficulty.

Q25. A gum bottle rests on its base. If it is placed upside

down, how does the (i) thrust (ii) pressure change?

(i)The effect of thrust depends on the area of the surface

on which it acts. When the gum bottle is placed upside

down, the surface area decreases as the bottle will be

now kept on its neck. Since the area decreases, the thrust

increases.

(ii) A gum bottle has narrow neck and wider base when

placed upside down. Since the area of contact decreases,

the pressure will increase.

Q26. Explain the following.

(a) Sleepers are used below the rails

Concrete sleepers (or wooden sleepers) arekept below

the iron rails of railway track so that the weightof

passing train is spread over a large area of ground

andthe pressure exerted by the rails on the ground

becomes less. This prevents the track to sink into the

ground.

(b) A tall building has wide foundation.

A tall building has wide foundation so that the weight

of the building may act on a larger area. As a result, it

will exert less pressure on the ground. This avoids

sinking of buildings into the earth.

Q27. Describe an experiment to show that a liquid exerts

pressure at the bottom of the container in which it is

kept.

Activity 3:- pg. no. 47. Diagram fig 3.21both (a) and (b)

Q28. Describe a suitable experiment to demonstrate that

a liquid exerts pressure sideways also.

Activity no.4,pg. no.47. Diagram 3.22 both (a) and (b)

Q29. Describe a simple experiment to show thatat a

given depth; a liquid exerts same pressure in all

directions.

Activity no.5,pg. no. 47 and 48. Diagram fig no.3.23

Q30. State two factors on which the pressure at a point in

a liquid depends.

The pressure at a point in a liquid depends on:

(a) The height of the liquid column. Liquid pressure

increases with the height of the liquid column

above the point

(b) The density of the liquid. Liquid pressure

increases with the increase in the density of the

liquid.

Q31. Describe an experiment to show that the liquid

pressure at a point increases with the increase in the

height of the liquid column above the point.

Activity 7,Pg no. 48. Diagram Fig no. 3.25

Q33. Describe an experiment to show that liquid pressure

depends on the density of the liquid.

Activity 8, pg. no. 49. Diagram Fig 3.26

Q34. A dam has broader walls at the bottom than at the

top. Give a reason.

A dam has broader walls at the bottom than at the top.

The reason is that the pressure at a point due to a liquid

increases with the increase in height of the liquid column

above it, so thickness of the walls of a dam is increased

towards the bottom so as to withstand the increasing

pressure of water.

Q35. What do you mean by atmospheric pressure?

Air has weight due to which it exerts a thrust on the

earth. The thrust on unit area of the earth surface due to

the column of air is called the atmospheric pressure. This

is about 105 Nm-2.

Q36. Write the numerical value of the atmospheric

pressure on the earth surface in pascal.

The atmospheric pressure is generally expressed in a unit

atmosphere (atm).

1 atm = 1.013 x 105 Pa

Q37. We do not feel uneasy even under the enormous

atmospheric pressure. Give a reason.

The surface area of an average human body is 2 m2.

Therefore a total thrust of about 2 x 105 N acts on our

body by the atmosphere. However we do not feel uneasy

under this enormous atmospheric pressure since the

blood in the veins of our body also exerts a pressure

(called the blood pressure) which is slightly more than

the atmospheric pressure. This blood pressure makes the

effect of atmospheric pressure ineffective.

Q38. Describe a simple experiment to illustrate that air

exerts pressure.

Activity No.9, pg. no. 50. Diagram Fig no. 3.29

Q39. Describe the crushing can experiment. What do we

conclude from this experiment?

Activity no. 10, pg. no. 60 Diagram 3.30 both (a) and (b)

Q40. Give reason for the following.

(a) A balloon collapses when air is removed from it.

A balloon bulges when air is filled into it due to the

pressure exerted by air which is blown into it. But it

collapses when air is removed from it. The reason is that

the outside atmospheric pressure exerted on the balloon

becomes greater than the pressure of air inside it. This

causes the balloon to collapse.

(b) Water does not run out from a dropper unless its

rubber bulb is pressed.

Atmospheric pressure acting from outside the dropper

balances the pressure exerted by water inside it and

water does not come out of a dropper. On pressing the

dropper, the inside pressure of water becomes greater

than the outside atmospheric pressure and thus water

runs out.

(c) Two holes are made in a sealed oil tin to take oil

out from it.

It is difficult to take out oil from a sealed tin if only one

hole is made in it. But if another hole is also made, the

atmospheric pressure acts on the oil due to air entering

in the tin through this hole and the oil then comes out of

the tin through the other hole easily.

Q41. How does atmospheric pressure change with

altitude?

As we rise up, the density of air decreases. Therefore

atmospheric pressure decreases with increasing altitude

I.e., as we go higher above the earth surface, the air

pressure decreases. Pressure is taken as 76 cm of

mercury column which is 1atm. For every 105 m rise in

height, the pressure decreases by 1cm of mercury

column.

NUMERICALS

1. Find the moment of force of 20 N about an axis of

rotation at a distance of 0.5 m from the force.

Force applied = 20 N.

Distance from the point of application of force = 0.5 m

To find: moment of force

Moment of force = Force x distance from the point of .

application of force.

= 20 N x 0.5 m = 10 Nm

Moment of force = 10 Nm

2. The moment of a force of 25N about a point is 2.5

Nm. Find the perpendicular distance of force from

that point.

Moment of force = 2.5 Nm

Force applied = 25 N

To find: perpendicular distance of force from the

point

Perpendicular distance= 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐹𝑜𝑟𝑐𝑒

𝐹𝑜𝑟𝑐𝑒 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 =

2.5 𝑁𝑚

25 𝑁

= 0.1 m = 10 cm

Distance of force = 10 cm

3. A spanner of length 10 cm is used to unscrew a nut

by applying a maximum force of 5.0N. Calculate the

moment of force required.

Length of the spanner = 10 cm = 0.1 m

Maximum Force applied = 5.0 N

To find: Moment of force required

Moment of force = Force x perpendicular distance

= 5.0 N x 0.1m = 0.5 Nm

Moment of force = 0.5 Nm

4. A wheel of diameter 2m can be rotated about an axis

passing through its centre by a moment of force

equal to 2.0 Nm. What minimum force must be

applied on its rim?

Perpendicular distance = 2m

Moment of force = 2.0 Nm

To find: minimum force applied

Force = 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒

𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =

2.0 𝑁𝑚

2𝑚 = 1N

Minimum force = 1N

5. A normal force of 200N acts on an area of 0.02 m2.

Find the pressure in pascal.

Force = 200 N

Area = 0.02 m2

To find: pressure exerted

Pressure = 𝐹𝑜𝑟𝑐𝑒

𝑎𝑟𝑒𝑎 =

200 𝑁

0.02 𝑚2 = 10000 N/m2

=10,000 pascal

Pressure exerted = 10,000 Pa

6. Find the thrust required to exert a pressure of

50,000pascal on an area of 0.05 m2.

Pressure exerted = 50,000pascal = 50,000 N/m2

Area = 0.05 m2

To find: thrust (force) required

Thrust (force) = Pressure x area

= 50,000 N/m2 x 0.05 m2

= 2500 N

Force = 2500 N

Q7. Find the area of a body which experiences a

pressure of 50,000 Pa by a thrust of 100N.

Force = 100 N

Pressure = 50,000 Pa = 50,000 N/m2

To find =area of the body

Area = 𝑓𝑜𝑟𝑐𝑒

𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒=

100 𝑁

50,000 𝑁/𝑚2 = 1

2000 m2

= 2 x 10-3 m2

Area of the body = 2 x 10-3 m2

Q8. Calculate the pressure in pascal exerted by a

force of 300 N acting normally on an area of 30 cm2.

Force = 300 N

Area = 30 cm2 = 30

10−4 m2 (1cm2 =

1

10−4 m2)

To find: pressure exerted

Pressure = 𝐹𝑜𝑟𝑐𝑒

𝑎𝑟𝑒𝑎 =

300 𝑁

30 𝑚2

104

= 300 𝑥 104

30 N/m2

= 10 x 104 N/m2 = 105 N/m2

= 105 Pa

Pressure exerted = 105 Pa

Q9. How much thrust will be required to exert a

pressure of 20,000 Pa on an area of 1 cm2?

Pressure = 20,000 Pa = 20,000 N/m2

Area = 1cm2 = 10-4m2

To find: Thrust (force) required

Force = Pressure x area

= 20,000 N/m2 x 10-4 m2 = 2 N

Force applied =2 N

Q10. The base of a container measures

15cmx 20 cm. It is placed on a table top. If the

weight of the container is 60 N, what is the pressure

exerted by the container on the table top?

Weight of the container (force) = 60 N

Area of the table top = 300cm2 = 300 x 10-4 m2

= 3x 10-2 m2

Pressure = 𝐹𝑜𝑟𝑐𝑒

𝑎𝑟𝑒𝑎 =

60 𝑁

3 x10−2𝑚2 =

60 x102

3 N/m2

=20 x 102 N/m2 =2000 N/m2

= 2000 Pa

Pressure = 2000 Pascal

Q11. Calculate the pressure exerted on a surface of

0.5 m2 by a thrust of 100kgf.

Thrust (force) = 100kgf

Area = 0.5 m2

To find: Pressure exerted

Pressure = 𝐹𝑜𝑟𝑐𝑒

𝑎𝑟𝑒𝑎=

100𝑘𝑔𝑓

0.5 𝑚2 = 200kgf m-2

Q12. A boy weighing 60kgf stands on a platform of

dimension 2.5 cm x 0.5 cm. what pressure in pascal

does he exert?

1kgf = 10 N

So 60kgf = 60 x 10 N = 600 N

Hence, force = 600 N

Area of surface = 2.5 cm x 0.5 cm = 1.25 cm2

(1cm2=10-4m2) =1.25 x 10-4 m2

To find: pressure exerted by the boy

Pressure = 𝐹𝑜𝑟𝑐𝑒

𝑎𝑟𝑒𝑎 =

600 𝑁

1.25 x10−4𝑚2 =

600 x104

1.25N/m2

=600 x104

125 x10−2 N/m2=

600 x102 x 104

125N/m2

600 x106

125 N/m2= 4.8 x 106 N/m2 = 4.8 x 106pascal

Pressure = 4.8 x 106 Pa

Q13. Fig 3.35 shows a brick of weight 2kgf and

dimensions 20 cm x 10 cm x 5 cm placed in three

different positions on the ground, Find the pressure

exerted by the brick in each case.

Case 1

Area = 20 cm x 10 cm =200 cm2

Force = 2kgf

Pressure = 𝐹𝑜𝑟𝑐𝑒

𝑎𝑟𝑒𝑎 =

2𝑘𝑔𝑓

200 𝑐𝑚2 = 0.01 kgfcm-2

Case2

Area = 5cm x 10 cm = 50 cm2

Force = 2kgf

Pressure = 𝐹𝑜𝑟𝑐𝑒

𝑎𝑟𝑒𝑎 =

2𝑘𝑔𝑓

50𝑐𝑚2 = 0.04 kgfcm-2

Case3

Area = 20cm x 5cm = 100 cm2

Force = 2kgf

Pressure = 𝐹𝑜𝑟𝑐𝑒

𝑎𝑟𝑒𝑎=

2𝑘𝑔𝑓

100𝑐𝑚2 = 0.02kgfcm-2

WORKSHEET

1. The moment of force of 10 N about point P is 50

Nm. Calculate the distance of point of application of

force about that point. (Ans- 5m)

2. Calculate the torque produced when a force of 50 N

acts on a body. The distance between point of

rotation and line of action of force is 40 cm.

(Ans- 20Nm)

3. Calculate the pressure produced by a force of 800 N

on an area of 20 cm2. (Ans - 4 x 105 N/m2)

4. A force of 100 N is applied over an object to produce

a pressure of 500 pascal. Calculate the area on which

force is applied.( Ans 0.2 m2)

Fill in the blanks.

1. Moment of force is ________ if it tends to rotate the

body clockwise.

2. The other name of moment of force is ________.

3. The force acting perpendicular on a unit area is

called ________.

4. As one goes up the mountains, the air pressure

________.

5. A ________ is used to measure liquid pressure.

6. The pressure in a liquid _______ with depth.

7. The empty space not containing air is called

________.

Answers

1. Negative 2. Torque 3. Thrust 4. Decreases

5. Manometer 6. Increases 7. vacuum